**College Algebra**

**Tutorial 40: ****Graphs of Rational Functions**

**Learning Objectives**

After completing this tutorial, you should be able to:

- Find the domain of a rational function.
- Find the vertical asymptote(s) of a rational function.
- Find the horizontal asymptote of a rational function.
- Find the oblique or slant asymptote of a rational function.
- Graph a rational function.

**Introduction**

In this tutorial we will be looking at several aspects
of rational functions. First we will revisit the concept of
domain. On rational functions, we need to be careful that we
don't use values of x that cause our denominator to be zero. If you need a review on
domain, feel free to go to Tutorial 30: Introductions to Functions.
Next, we look at vertical, horizontal and slant asymptotes.
Basically an asymptote is an imaginary line that the curve of
the function
gets very close to or approaches. In the end, we put it all
together and graph rational functions. Sounds like fun, you
better get to it!!!

** Tutorial**

**Review on Domain
**

The **domain **is the set of all **input values** to
which the
rule applies. These are called your **independent variables**.
These are the values that correspond to the first components of the
ordered
pairs it is associated with.

If you need a review on domain, feel free to go to Tutorial 30: Introductions to Functions.

If you need a review on domain, feel free to go to Tutorial 30: Introductions to Functions.

Our restriction here is that the denominator of a
fraction can never
be equal to 0. So to find our domain, we want to set the
denominator
equal to 0 and restrict those values.

Our domain is all real
numbers except 1/2 and -3, because 1/2 and -3 both make the
denominator equal to 0, which would not give us a real number answer
for our function.

**Let
be written in lowest terms and
P and Q **

**If or as , then
the vertical line x = a is a vertical asymptote.
**

**The line x = a is a vertical asymptote of the
graph of f
if and only if the denominator Q(a)
= 0
and the numerator .
**

In this case that is where the simplified rational function’s denominator is equal to 0.

**Some things to note:**

You can have zero or many vertical
asymptotes. It
will be *x* = whatever number(s) cause the denominator to be zero after you have
simplified
the function.

You draw a vertical asymptote on the graph by
putting a
dashed vertical
(up and down) line going through *x* = *a* as shown below. This is where the function is undefined, so there
will be NO point on the vertical asymptote itself. The graph will
approach
it from both sides, but never cross over it.

Below is an
example of a vertical asymptote of x = 2:

**First we want to check and see if this rational
function
will reduce
down**:

***Factor the function**

Nothing is able to cancel out, so now **we want to
find
where the
denominator is equal to 0:**

***Set the first factor = 0**

***Set the second factor = 0**

There are **two vertical asymptotes: ***x* = -3 and* x* = 3.

**First we want to check and see if this rational
function
will reduce
down**:

***Factor the function**

***Cancel out the common factor
of x + 1**

Note how the factor x + 1 canceled out, so now
we want to find where the new
denominator is equal to 0:

There is one** vertical asymptote: ***x* = -2.

**Let be written in lowest terms,
where P and Q are polynomial functions and .
**

**If as or , then
the horizontal line y = a is
a horizontal asymptote.**

Case I

If the degree of P(x) < the degree of Q(x), then there is a horizontal asymptote at y = 0 (x-axis).

Case II

If the degree of P(x) = the degree of Q(x), then there is a horizontal asymptote at

In other words, it would be the ratio between the leading coefficient of the numerator and the leading coefficient of the denominator.

**Some things to note:**

You may have 0 or 1 horizontal asymptote,
but no
more than that.

The graph may cross the horizontal asymptote, but it levels off and approaches it as*x* goes to infinity.

The graph may cross the horizontal asymptote, but it levels off and approaches it as

You draw a horizontal asymptote on the graph by
putting
a dashed horizontal
(left and right) line going through *y* = *a*.

Below is an example of a horizontal asymptote of y = 3:

**First we want to check and see if this rational
function
will reduce
down**:

***Factor the function**

Nothing was able to cancel out, so now we want to
compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 1, you are correct. The leading term is 3x and its degree is 1.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is 1 is less than the degree of the denominator, then there is a horizontal asymptote at y = 0.

What is the degree of the numerator? If you said 1, you are correct. The leading term is 3x and its degree is 1.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is 1 is less than the degree of the denominator, then there is a horizontal asymptote at y = 0.

**First we want to check and see if this rational
function
will reduce
down**:

***Factor the function**

***Cancel out the common factor
of x + 1**

Note how the factor x + 1 canceled out, so now we
want to compare the degrees of the numerator and the denominator that
is left.

What is the degree of the numerator that is left? If you said 1, you are correct. The leading term is x and its degree is 1.

What is the degree of the denominator that is left? If you said 1, you are correct. The leading term is x and its degree is 1.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at

What is the degree of the numerator that is left? If you said 1, you are correct. The leading term is x and its degree is 1.

What is the degree of the denominator that is left? If you said 1, you are correct. The leading term is x and its degree is 1.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at

that is neither vertical or horizontal.

If the degree of the numerator is one more than

the degree of the denominator, then the graph of

the rational function will have a slant asymptote.

**Some things to note:**

The slant asymptote is the quotient part of the answer you get when you
divide the numerator by the denominator.

If you need a review of long division, feel free to go to Tutorial 36: Long Division.

You may have 0 or 1 slant asymptote, but no more than that.

A graph can have both a vertical and a slant asymptote, but it CANNOT have both a horizontal and slant asymptote.

You draw a slant asymptote on the graph by
putting
a dashed horizontal
(left and right) line going through *y* = mx + b.

Below is an
example of a slant asymptote of y = x + 1:

**Note that this rational function is already reduced
down.
**

Applying long
division to this problem we get:

The answer to the long division would be .

The equation for the slant asymptote is the quotient part of the answer which would be .

The equation for the slant asymptote is the quotient part of the answer which would be .

If any factors are TOTALLY removed from the
denominator, then
there will not be a vertical asymptote through that value, but an open
hole at that point.

If this is the case, plug in the x value that causes that removed factor to be zero into the reduced rational function. Plot this point as an open hole.

If this is the case, plug in the x value that causes that removed factor to be zero into the reduced rational function. Plot this point as an open hole.

Let be a rational function
reduced to lowest terms and Q(x)
has a degree of at least 1:

There is a vertical asymptote for every
root of .

There is a horizontal asymptote of y = 0 (x-axis) if the degree of P(x) < the degree of Q(x).

There is a horizontal asymptote of

if the degree of P(x) = the degree of Q(x).

There is an oblique or slant asymptote if the degree of P(x) is one degree higher than Q(x). If this is the case the oblique asymptote is the quotient part of the division.

There is a horizontal asymptote of y = 0 (x-axis) if the degree of P(x) < the degree of Q(x).

There is a horizontal asymptote of

if the degree of P(x) = the degree of Q(x).

There is an oblique or slant asymptote if the degree of P(x) is one degree higher than Q(x). If this is the case the oblique asymptote is the quotient part of the division.

Note that a graph can have both a vertical and a slant asymptote, or both a vertical and horizontal asymptote, but it CANNOT have both a horizontal and slant asymptote.

The graph is **symmetric about
the y-axis** if the function is **even**.

The graph is**symmetric about the origin** if the function is **odd**.

If you need a review of even and odd functions, feel free to go to Tutorial 32: Graphs of Functions Part II.

The graph is

If you need a review of even and odd functions, feel free to go to Tutorial 32: Graphs of Functions Part II.

The x-intercept is
where the graph crosses the x-axis.
You can find this by setting y = 0 and solving for x.

The y-intercept is where the graph crosses the y-axis. You can find this by setting x = 0 and solving for y.

If you need a review on intercepts, feel free to go to Tutorial 26: Equations of Lines.

The y-intercept is where the graph crosses the y-axis. You can find this by setting x = 0 and solving for y.

If you need a review on intercepts, feel free to go to Tutorial 26: Equations of Lines.

You should have AT LEAST two points in each section of
the graph that is marked off by the vertical asymptotes.

Note that your graph can cross over a horizontal or
oblique asymptote, but it can NEVER cross over a vertical asymptote.

This function cannot be reduced any further. This
means that there will be no open holes on this graph.

Vertical Asymptote:

So now we want to find where the denominator is equal to 0:

So now we want to find where the denominator is equal to 0:

There is one** vertical asymptote: ***x* = 0.

Horizontal Asymptote:

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 0, you are correct. The leading term is the constant -1 and its degree is 0.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is less than the degree of the denominator, then there is a horizontal asymptote at y = 0.

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 0, you are correct. The leading term is the constant -1 and its degree is 0.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is less than the degree of the denominator, then there is a horizontal asymptote at y = 0.

Slant Asymptote:

Since the degree of the numerator is NOT one degree higher than the degree of the denominator, there is not slant asymptote.

Since the degree of the numerator is NOT one degree higher than the degree of the denominator, there is not slant asymptote.

Since , the function is even.

This means the graph is symmetric about the y-axis.

If you need a review of even and odd functions, feel free to go to Tutorial 32: Graphs of Functions Part II.

x-intercept:

What value are we going to use for y? You are correct if you said 0.

What value are we going to use for y? You are correct if you said 0.

This means there is NO x-intercept.

y-intercept:

What value are we going to use for x? You are correct if you said 0.

What value are we going to use for x? You are correct if you said 0.

This means there is NO y-intercept.

So far we have not found any points to plot on the
graph. Note how the vertical asymptote sections the graph into
two parts. I’m going to plug in two x values that are to the left
of x = 0 and two
that are to the right of x = 0.

Plugging in -2 for x we get:

(-2, -1/4)

(-2, -1/4)

Plugging in -1 for x we get:

(-1, -1)

(-1, -1)

Plugging in 1 for x we get:

(1, -1)

(1, -1)

Plugging in 2 for x we get:

(2, -1/4)

(2, -1/4)

This function cannot be reduced any further. This
means that there will be no open holes on this graph.

Vertical Asymptote:

So now we want to find where the denominator is equal to 0:

So now we want to find where the denominator is equal to 0:

There are two** vertical asymptotes: ***x* = -2 and x = 1.

Horizontal Asymptote:

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 2, you are correct. The leading term is and its degree is 2.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at

So now we want to compare the degrees of the numerator and the denominator.

What is the degree of the numerator? If you said 2, you are correct. The leading term is and its degree is 2.

What is the degree of the denominator? If you said 2, you are correct. The leading term is and its degree is 2.

If you need a review of finding the degree of a polynomial, feel free to go to Tutorial 6: Polynomials.

Since the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at

Slant Asymptote:

Since the degree of the numerator is NOT one degree higher than the degree of the denominator, there is not slant asymptote.

Since the degree of the numerator is NOT one degree higher than the degree of the denominator, there is not slant asymptote.

Since , the function is neither even nor odd.

If you need a review of even and odd functions, feel free to go to Tutorial 32: Graphs of Functions Part II.

x-intercept:

What value are we going to use for y? You are correct if you said 0.

What value are we going to use for y? You are correct if you said 0.

This means there is NO x-intercept.

y-intercept:

What value are we going to use for x? You are correct if you said 0.

What value are we going to use for x? You are correct if you said 0.

The y-intercept is (0, -1/2)

Note how the
vertical asymptotes section the graph into three parts. I’m going
to
plug in two x values that are to the left of x = -2, one in between x = -2 and x = 1,
and two that are to the right of x = 1.

Plugging in -4 for x we get:

(-4, 17/10)

(-4, 17/10)

Plugging in -3 for x we get:

(-3, 5/2)

(-3, 5/2)

Plugging in -1 for x we get:

(-1, -1)

(-1, -1)

Plugging in 2 for x we get:

(2, 3/4)

(2, 3/4)

Plugging in 3 for x we get:

(3, 1)

(3, 1)

The factor of x canceled out and there were no factors of *x* left in
the denominator. This means there is an open hole on the graph at x = 0.

x = 0:

x = 0:

There is an open hole
at (0, -4/3).

Vertical Asymptote:

So now we want to find where the new denominator is equal to 0:

So now we want to find where the new denominator is equal to 0:

There is one** vertical asymptote: ***x* = -3.

Horizontal Asymptote:

Since the degree of the numerator is one degree higher than the degree of the denominator, there is a slant asymptote and no horizontal asymptote.

Since the degree of the numerator is one degree higher than the degree of the denominator, there is a slant asymptote and no horizontal asymptote.

Slant Asymptote:

**Applying long
division to this problem we get:**

The answer to the long division would be .

The equation for the slant asymptote is the quotient part of the answer which would be .

The equation for the slant asymptote is the quotient part of the answer which would be .

Since , the function is neither even nor odd.

If you need a review of even and odd functions, feel free to go to Tutorial 32: Graphs of Functions Part II.

x-intercept:

What value are we going to use for y? You are correct if you said 0.

What value are we going to use for y? You are correct if you said 0.

There are two x-intercepts: (4, 0) and (-1,
0).

y-intercept:

What value are we going to use for x? You are correct if you said 0.

What value are we going to use for x? You are correct if you said 0.

The y-intercept is (0, -4/3).

Note that this will be open hole, as found in step 1.

Note that this will be open hole, as found in step 1.

Note how the
vertical asymptote sections the graph into two parts. I’m going
to
plug in two x values that are to the left of x = -3.

Plugging in -5 for x we get:

(-5, -18)

(-5, -18)

Plugging in -4 for x we get:

(-4, -24)

(-4, -24)

** Practice Problems**

These are practice problems to help bring you to the next level.
It will allow you to check and see if you have an understanding of these
types of problems. **Math works just like anything
else, if you want to get good at it, then you need to practice it.
Even the best athletes and musicians had help along the way and lots of
practice, practice, practice, to get good at their sport or instrument.**
In fact there is no such thing as too much practice.

To get the most out of these, **you should work the problem out on
your own and then check your answer by clicking on the link for the answer/discussion
for that problem**. At the link you will find the answer
as well as any steps that went into finding that answer.

Practice Problem 1a:Give the domain of the given function.

Practice Problems 2a - 2b: Find the vertical and horizontal asymptotes for the given functions.

Practice Problem 3a:Find the oblique asymptote for the given function.

Practice Problems 4a - 4b: Sketch the graph of the given function.

** Need Extra Help on these Topics?**

**The following are webpages that can assist you in the topics that were covered on this page**.

This website helps you with graphing rational functions.

http://www.purplemath.com/modules/grphrtnl2.htm

This website helps you with graphing rational functions.

http://www.purplemath.com/modules/grphrtnl3.htm

This website helps you with graphing rational functions.

**Go to Get
Help Outside the
Classroom found in Tutorial 1: How to Succeed in a Math Class for
some
more suggestions.**

Last revised on March 18, 2011 by Kim Seward.

All contents copyright (C) 2002 - 2011, WTAMU and Kim Seward. All rights reserved.